Two particles are performing SHM with same amplitude and time period. At an instant two particles are having velocity `1m//s` but one is on the right and the other is on left of their mean positiion. When the particles have same position there speed is `sqrt(3)m//s`. Find the maximum speed (in m/s) of particles during SHM.

Explanation

To solve the problem, we need to find the maximum speed of two particles performing Simple Harmonic Motion (SHM) given their velocities at specific positions.

### Step-by-Step Solution:

1. **Understanding the SHM Equations**:
The displacement of a particle in SHM can be expressed as:
\[ x = A \sin(\omega t) \]
The velocity of the particle can be derived from the displacement:
\[ v = \frac{dx}{dt} = \omega A \cos(\omega t) \]
Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is time.

2. **Given Conditions**:
- At a certain instant, both particles have a velocity of \( 1 \, \text{m/s} \) but are at opposite sides of the mean position.
- When the particles are at the same position, their speed is \( \sqrt{3} \, \text{m/s} \).

3. **Setting Up the Equations**:
Let \( V_{\text{max}} \) be the maximum speed of the particles during SHM.
- For the first position (where one particle is on the right and the other on the left):
\[ 1 = V_{\text{max}} \cos(\theta) \]
- For the second position (when both particles have the same position):
\[ \sqrt{3} = V_{\text{max}} \cos(\phi) \]

4. **Relating the Angles**:
The angles \( \theta \) and \( \phi \) correspond to the positions of the particles. Since they are at opposite sides initially, we can relate them:
\[ \phi = \frac{\pi}{2} + \theta \]
Thus, we can express \( \cos(\phi) \) as:
\[ \cos(\phi) = \cos\left(\frac{\pi}{2} + \theta\right) = -\sin(\theta) \]

5. **Substituting into the Equations**:
From the first equation:
\[ V_{\text{max}} = \frac{1}{\cos(\theta)} \]
From the second equation:
\[ \sqrt{3} = V_{\text{max}} (-\sin(\theta)) \]
Substituting \( V_{\text{max}} \) from the first equation into the second:
\[ \sqrt{3} = -\frac{1}{\cos(\theta)} \sin(\theta) \]
Rearranging gives:
\[ \sqrt{3} \cos(\theta) = -\sin(\theta) \]

6. **Using Trigonometric Identity**:
Dividing both sides by \( \cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)):
\[ \sqrt{3} = -\tan(\theta) \]
This implies:
\[ \tan(\theta) = -\sqrt{3} \]
Therefore, \( \theta = \frac{5\pi}{3} \) or \( \theta = \frac{2\pi}{3} \).

7. **Finding Maximum Speed**:
Now, substituting \( \theta \) back into the equation for \( V_{\text{max}} \):
\[ V_{\text{max}} = \frac{1}{\cos(\theta)} \]
We can find \( \cos(\theta) \) using \( \tan(\theta) = -\sqrt{3} \):
\[ \cos(\theta) = \frac{1}{2} \quad \text{(since } \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\text{)} \]
Therefore:
\[ V_{\text{max}} = \frac{1}{\frac{1}{2}} = 2 \, \text{m/s} \]

### Final Answer:
The maximum speed of the particles during SHM is \( 2 \, \text{m/s} \).